[proofplan] The argument is the standard Hilbert-space orthogonality argument for a self-adjoint operator. First we use self-adjointness to show that the two eigenvalues are real, because each eigenform has positive Petersson norm. Then we compute $(T_n f,g)_{\mathrm{Pet}}$ in two ways: once from the eigenvalue equation for $f$, and once from self-adjointness and the eigenvalue equation for $g$. The resulting identity $(\lambda-\mu)(f,g)_{\mathrm{Pet}}=0$ forces orthogonality when $\lambda \ne \mu$. [/proofplan]

[step:Use self-adjointness to show that the eigenvalues are real]Because $f \ne 0$ and the Petersson inner product is positive definite on $S_k(SL_2(\mathbb{Z}))$, we have $(f,f)_{\mathrm{Pet}} > 0$. Using $T_n f = \lambda f$, self-adjointness of $T_n$, and conjugate-linearity in the second argument, \begin{align*} \lambda (f,f)_{\mathrm{Pet}} &= (T_n f,f)_{\mathrm{Pet}} \\ &= (f,T_n f)_{\mathrm{Pet}} \\ &= (f,\lambda f)_{\mathrm{Pet}} \\ &= \overline{\lambda}(f,f)_{\mathrm{Pet}}. \end{align*} Since $(f,f)_{\mathrm{Pet}} > 0$, division gives $\lambda=\overline{\lambda}$, so $\lambda \in \mathbb{R}$. The same computation with $g$ gives \begin{align*} \mu (g,g)_{\mathrm{Pet}} &= (T_n g,g)_{\mathrm{Pet}} \\ &= (g,T_n g)_{\mathrm{Pet}} \\ &= (g,\mu g)_{\mathrm{Pet}} \\ &= \overline{\mu}(g,g)_{\mathrm{Pet}}. \end{align*} Because $g \ne 0$, $(g,g)_{\mathrm{Pet}} > 0$, hence $\mu=\overline{\mu}$ and $\mu \in \mathbb{R}$.[/step]

[guided]We first record that the eigenvalues are real. This is not a separate hypothesis: it follows from self-adjointness. Since $f$ is a nonzero cusp form and $(\cdot,\cdot)_{\mathrm{Pet}}$ is an inner product, its Petersson norm is positive: \begin{align*} (f,f)_{\mathrm{Pet}} > 0. \end{align*} Now compute the same scalar $(T_n f,f)_{\mathrm{Pet}}$ in two ways. The eigenvalue equation $T_n f=\lambda f$ gives \begin{align*} (T_n f,f)_{\mathrm{Pet}}=\lambda(f,f)_{\mathrm{Pet}}, \end{align*} because the Petersson inner product is linear in the first argument. Self-adjointness gives \begin{align*} (T_n f,f)_{\mathrm{Pet}}=(f,T_n f)_{\mathrm{Pet}}. \end{align*} Substituting $T_n f=\lambda f$ into the second argument and using conjugate-linearity in that argument gives \begin{align*} (f,T_n f)_{\mathrm{Pet}}=(f,\lambda f)_{\mathrm{Pet}}=\overline{\lambda}(f,f)_{\mathrm{Pet}}. \end{align*} Combining these identities, \begin{align*} \lambda(f,f)_{\mathrm{Pet}}=\overline{\lambda}(f,f)_{\mathrm{Pet}}. \end{align*} The factor $(f,f)_{\mathrm{Pet}}$ is strictly positive, so it can be divided out. Hence $\lambda=\overline{\lambda}$, which means $\lambda \in \mathbb{R}$. The same argument applies to $g$. Since $g \ne 0$, we have $(g,g)_{\mathrm{Pet}}>0$, and \begin{align*} \mu (g,g)_{\mathrm{Pet}} &= (T_n g,g)_{\mathrm{Pet}} \\ &= (g,T_n g)_{\mathrm{Pet}} \\ &= (g,\mu g)_{\mathrm{Pet}} \\ &= \overline{\mu}(g,g)_{\mathrm{Pet}}. \end{align*} Dividing by $(g,g)_{\mathrm{Pet}}>0$ gives $\mu=\overline{\mu}$, so $\mu \in \mathbb{R}$.[/guided]

[step:Compare the two evaluations of $(T_n f,g)_{\mathrm{Pet}}$]Using the eigenvalue equation $T_n f=\lambda f$ and linearity in the first argument, \begin{align*} (T_n f,g)_{\mathrm{Pet}}=\lambda(f,g)_{\mathrm{Pet}}. \end{align*} Using self-adjointness, the eigenvalue equation $T_n g=\mu g$, and conjugate-linearity in the second argument, \begin{align*} (T_n f,g)_{\mathrm{Pet}} &= (f,T_n g)_{\mathrm{Pet}} \\ &= (f,\mu g)_{\mathrm{Pet}} \\ &= \overline{\mu}(f,g)_{\mathrm{Pet}}. \end{align*} Since $\mu \in \mathbb{R}$, $\overline{\mu}=\mu$. Therefore \begin{align*} \lambda(f,g)_{\mathrm{Pet}}=\mu(f,g)_{\mathrm{Pet}}. \end{align*} Subtracting the right-hand side from the left-hand side gives \begin{align*} (\lambda-\mu)(f,g)_{\mathrm{Pet}}=0. \end{align*}[/step]

[guided]We now compute $(T_n f,g)_{\mathrm{Pet}}$ in two different ways. The first computation uses the eigenvalue equation for $f$: \begin{align*} (T_n f,g)_{\mathrm{Pet}}=(\lambda f,g)_{\mathrm{Pet}}=\lambda(f,g)_{\mathrm{Pet}}, \end{align*} where the scalar $\lambda$ leaves unchanged because the Petersson inner product is linear in the first argument. The second computation moves $T_n$ from the first slot to the second slot. This is exactly the self-adjointness hypothesis: \begin{align*} (T_n f,g)_{\mathrm{Pet}}=(f,T_n g)_{\mathrm{Pet}}. \end{align*} Now use the eigenvalue equation $T_n g=\mu g$: \begin{align*} (f,T_n g)_{\mathrm{Pet}}=(f,\mu g)_{\mathrm{Pet}}. \end{align*} Because the Petersson inner product is conjugate-linear in the second argument, \begin{align*} (f,\mu g)_{\mathrm{Pet}}=\overline{\mu}(f,g)_{\mathrm{Pet}}. \end{align*} From the previous step, $\mu$ is real, so $\overline{\mu}=\mu$. Thus the second computation gives \begin{align*} (T_n f,g)_{\mathrm{Pet}}=\mu(f,g)_{\mathrm{Pet}}. \end{align*} Comparing the first and second computations, \begin{align*} \lambda(f,g)_{\mathrm{Pet}}=\mu(f,g)_{\mathrm{Pet}}. \end{align*} After subtracting the right-hand side, we obtain \begin{align*} (\lambda-\mu)(f,g)_{\mathrm{Pet}}=0. \end{align*}[/guided]

[step:Use the distinctness of the eigenvalues to force orthogonality] By hypothesis $\lambda \ne \mu$, so $\lambda-\mu \ne 0$. Since \begin{align*} (\lambda-\mu)(f,g)_{\mathrm{Pet}}=0, \end{align*} division by the nonzero scalar $\lambda-\mu$ gives \begin{align*} (f,g)_{\mathrm{Pet}}=0. \end{align*} This is exactly the asserted Petersson orthogonality of $f$ and $g$. [/step]